
If a capillary tube has been dipped in a liquid, the liquid will rise to a height $h$ in the tube. It has been given that the free liquid surface inside the tube will be hemispherical in the shape. The tube will be now pushed down so that the height of the tube outside the liquid will be less than $h$. Then,
This question will be having multiple correct options
A. The liquid has come out of the tube like in a small fountain.
B. The liquid will be oozing out of the tube slowly.
C. The liquid will fill the tube but not come out of its upper end.
D. The free liquid surface will not be hemispherical inside the tube.
Answer
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Hint: The height to which the liquid will be rising can be found by taking the ratio of twice the product of the surface tension of the liquid and the cosine of the angle to the product of the radius of the liquid meniscus, the density and the acceleration due to gravity. This will help you in answering this question.
Complete answer:
First of all let us consider a capillary tube of radius $r$ which will be immersed in a liquid of surface tension $T$ and the density $\rho $. The height to which the liquid will be rising in the capillary tube has been given as,
$h=\dfrac{2T\cos \theta }{R\rho g}$
Where $T$ be the surface tension of the liquid, $R$ be the liquid meniscus, $\rho $ be the density and $g$ be the acceleration due to gravity. The angle will be perpendicular.
$h=\dfrac{2T\cos \theta }{R\rho g}=\dfrac{2T}{R\rho g}$
Rearranging this equation can be written as,
$hR=\dfrac{2T}{\rho g}$
This has been found to be a constant.
${{h}_{1}}{{R}_{1}}={{h}_{2}}{{R}_{2}}$
When the tube has been pushed down we are enhancing the $h$, thereby decreasing the radius $R$ of the liquid meniscus. Therefore as $h$ is increased, the level of the liquid becomes more and more flat but will not get overflowed. The angle of contact at the free liquid surface inside the capillary tube will be varying such that the vertical component of the surface tension forces will be balancing the weight of the liquid column.
Therefore the correct answers have been mentioned as option C and D.
Note:
Capillary action can be otherwise called as capillarity, capillary effect or wicking. This can be defined as the ability of a liquid to flow in the narrow spaces without the assistance of, or even in prevention to, external forces like gravity. It will be happening due to the intermolecular between the liquid and surrounding solid surfaces.
Complete answer:
First of all let us consider a capillary tube of radius $r$ which will be immersed in a liquid of surface tension $T$ and the density $\rho $. The height to which the liquid will be rising in the capillary tube has been given as,
$h=\dfrac{2T\cos \theta }{R\rho g}$
Where $T$ be the surface tension of the liquid, $R$ be the liquid meniscus, $\rho $ be the density and $g$ be the acceleration due to gravity. The angle will be perpendicular.
$h=\dfrac{2T\cos \theta }{R\rho g}=\dfrac{2T}{R\rho g}$
Rearranging this equation can be written as,
$hR=\dfrac{2T}{\rho g}$
This has been found to be a constant.
${{h}_{1}}{{R}_{1}}={{h}_{2}}{{R}_{2}}$
When the tube has been pushed down we are enhancing the $h$, thereby decreasing the radius $R$ of the liquid meniscus. Therefore as $h$ is increased, the level of the liquid becomes more and more flat but will not get overflowed. The angle of contact at the free liquid surface inside the capillary tube will be varying such that the vertical component of the surface tension forces will be balancing the weight of the liquid column.
Therefore the correct answers have been mentioned as option C and D.
Note:
Capillary action can be otherwise called as capillarity, capillary effect or wicking. This can be defined as the ability of a liquid to flow in the narrow spaces without the assistance of, or even in prevention to, external forces like gravity. It will be happening due to the intermolecular between the liquid and surrounding solid surfaces.
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